仅由元音组成的子串计数

给定字符串S ，任务是计算仅包含元音的所有子字符串。

例子：

Input: S = “geeksforgeeks” 
Output: 7 
Explanation: 
Substrings {“e”, “ee”, “e”, “o”, “e”, “ee”, “e”} consists only of vowels.

Input: S = “aecui” 
Output: 6 
Explanation: 
Substrings {“a”, “ae”, “e”, “u”, “ui”, “i”} consists only of vowels. 

天真的方法：
最简单的方法是生成所有子字符串，并检查每个子字符串是否仅包含元音。

下面是上述方法的实现：

C++

// C++ program to Count all substrings
// in a string which contains only vowels
#include 
using namespace std;
 
// Function to check if a
// character is vowel or not
bool isvowel(char ch)
{
    return (ch == 'a' or ch == 'e'
            or ch == 'i' or ch == 'o'
            or ch == 'u');
}
 
// Function to check whether
// string contains only vowel
bool isvalid(string& s)
{
    int n = s.length();
 
    for (int i = 0; i < n; i++) {
        // Check if the character is
        // not vowel then invalid
        if (!isvowel(s[i]))
            return false;
    }
 
    return true;
}
 
// Function to Count all substrings
// in a string which contains
// only vowels
int CountTotal(string& s)
{
    int ans = 0;
 
    int n = s.length();
 
    // Generate all substring of s
    for (int i = 0; i < n; i++) {
        string temp = "";
        for (int j = i; j < n; j++) {
            temp += s[j];
 
            // If temp contains only vowels
            if (isvalid(temp))
                // Increment the count
                ans += 1;
        }
    }
 
    return ans;
}
 
// Driver Program
int main()
{
    string s = "aeoibsddaaeiouudb";
 
    cout << (CountTotal(s)) << endl;
 
    return 0;
}

Java

// Java program to count all subStrings
// in a String which contains only vowels
import java.util.*;
 
class GFG{
 
// Function to check if a
// character is vowel or not
static boolean isvowel(char ch)
{
    return (ch == 'a' || ch == 'e' ||
            ch == 'i' || ch == 'o' ||
            ch == 'u');
}
 
// Function to check whether
// String contains only vowel
static boolean isvalid(String s)
{
    int n = s.length();
 
    for(int i = 0; i < n; i++)
    {
         
       // Check if the character is
       // not vowel then invalid
       if (!isvowel(s.charAt(i)))
           return false;
    }
    return true;
}
 
// Function to Count all subStrings
// in a String which contains
// only vowels
static int CountTotal(String s)
{
    int ans = 0;
    int n = s.length();
 
    // Generate all subString of s
    for(int i = 0; i < n; i++)
    {
       String temp = "";
       for(int j = i; j < n; j++)
       {
          temp += s.charAt(j);
           
          // If temp contains only vowels
          if (isvalid(temp))
               
              // Increment the count
              ans += 1;
       }
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    String s = "aeoibsddaaeiouudb";
 
    System.out.print((CountTotal(s)) + "\n");
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to count all substrings
# in a string which contains only vowels
 
# Function to check if a
# character is vowel or not
def isvowel(ch):
    return (ch == 'a' or ch == 'e' or
            ch == 'i' or ch == 'o' or
            ch == 'u')
 
# Function to check whether
# string contains only vowel
def isvalid(s):
     
    n = len(s)
    for i in range(n):
         
        # Check if the character is
        # not vowel then invalid
        if (not isvowel(s[i])):
            return False
             
    return True
 
# Function to Count all substrings
# in a string which contains
# only vowels
def CountTotal(s):
     
    ans = 0
    n = len(s)
 
    # Generate all substring of s
    for i in range(n):
        temp = ""
         
        for j in range(i, n):
            temp += s[j]
 
            # If temp contains only vowels
            if (isvalid(temp)):
                 
                # Increment the count
                ans += 1
                 
    return ans
 
# Driver code
if __name__ == '__main__':
     
    s = "aeoibsddaaeiouudb"
 
    print(CountTotal(s))
 
# This code is contributed by mohit kumar 29

C#

// C# program to count all subStrings
// in a String which contains only vowels
using System;
class GFG{
 
// Function to check if a
// character is vowel or not
static Boolean isvowel(char ch)
{
    return (ch == 'a' || ch == 'e' ||
            ch == 'i' || ch == 'o' ||
            ch == 'u');
}
 
// Function to check whether
// String contains only vowel
static Boolean isvalid(string s)
{
    int n = s.Length;
 
    for(int i = 0; i < n; i++)
    {
         
        // Check if the character is
        // not vowel then invalid
        if (!isvowel(s[i]))
            return false;
    }
    return true;
}
 
// Function to Count all subStrings
// in a String which contains
// only vowels
static int CountTotal(string s)
{
    int ans = 0;
    int n = s.Length;
 
    // Generate all subString of s
    for(int i = 0; i < n; i++)
    {
        string temp = "";
        for(int j = i; j < n; j++)
        {
            temp += s[j];
             
            // If temp contains only vowels
            if (isvalid(temp))
                 
                // Increment the count
                ans += 1;
        }
    }
    return ans;
}
 
// Driver code
public static void Main()
{
    string s = "aeoibsddaaeiouudb";
 
    Console.Write((CountTotal(s)) + "\n");
}
}
 
// This code is contributed by Code_Mech

Javascript

 
// Javascript program to count all subStrings
// in a String which contains only vowels
 
// Function to check if a
// character is vowel or not
function isvowel(ch)
{
    return (ch == 'a' || ch == 'e' ||
            ch == 'i' || ch == 'o' ||
            ch == 'u');
}
 
// Function to check whether
// String contains only vowel
function isvalid(s)
{
    let n = s.length;
 
    for(let i = 0; i < n; i++)
    {
 
        // Check if the character is
        // not vowel then invalid
        if (!isvowel(s[i]))
            return false;
    }
    return true;
}
 
// Function to Count all subStrings
// in a String which contains
// only vowels
function CountTotal(s)
{
    let ans = 0;
    let n = s.length;
 
    // Generate all subString of s
    for(let i = 0; i < n; i++)
    {
        let temp = "";
        for(let j = i; j < n; j++)
        {
            temp += s[j];
 
            // If temp contains only vowels
            if (isvalid(temp))
 
                // Increment the count
                ans += 1;
        }
    }
    return ans;
}
 
// Driver code
let s = "aeoibsddaaeiouudb";
 
document.write((CountTotal(s)));
 
// This code is contributed by divyesh072019
 

输出：

38

时间复杂度： O(N ³ )

高效方法：
为了优化上述方法，主要思想是计算仅包含元音(例如x)的子串的长度。然后，对于每个x ，可能的子字符串数为x *(x + 1)/ 2 ，其中仅包含元音。对每个这样的子字符串重复此过程，并返回最终计数。

下面是上述方法的实现：

C++

// C++ program to Count all substrings
// in a string which contains only vowels
#include 
using namespace std;
 
// Function to check vowel or not
bool isvowel(char ch)
{
    return (ch == 'a' or ch == 'e'
            or ch == 'i' or ch == 'o'
            or ch == 'u');
}
 
// Function to Count all substrings
// in a string which contains
// only vowels
int CountTotal(string& s)
{
    int ans = 0;
    int n = s.length();
    int cnt = 0;
 
    for (int i = 0; i < n; i++) {
        // Check if current character
        // is vowel
        if (isvowel(s[i]))
            // Increment length of substring
            cnt += 1;
 
        else {
            // Calculate possible
            // substrings of calculated
            // length
            ans += (cnt * (cnt + 1) / 2);
 
            // Reset the length
            cnt = 0;
        }
    }
 
    // Add remaining possible
    // substrings consisting
    // of vowels occupying
    // last indices of the string
    if (cnt != 0) {
        ans += (cnt * (cnt + 1) / 2);
    }
 
    return ans;
}
 
// Driver Program
int main()
{
    string s = "geeksforgeeks";
 
    cout << (CountTotal(s)) << endl;
 
    return 0;
}

Java

// Java program to Count all substrings
// in a string which contains only vowels
import java.io.*;
 
public class GFG {
 
    // Function to check vowel or not
    static boolean isvowel(char x)
    {
        return (x == 'a' || x == 'e'
                || x == 'i' || x == 'o'
                || x == 'u');
    }
 
    // Function to find largest
    // string which satisy condition
    static int CountTotal(String str)
    {
        int ans = 0;
 
        int n = str.length();
 
        char[] s = str.toCharArray();
 
        int cnt = 0;
 
        for (int i = 0; i < n; i++) {
            // Check if current character
            // is vowel
            if (isvowel(s[i]))
                // Increment count
                cnt += 1;
 
            else {
                // Count all possible
                // substrings of calculated
                // length
                ans += (cnt * (cnt + 1) / 2);
 
                // Reset the length
                cnt = 0;
            }
        }
 
        // Add remaining possible
        // substrings consisting
        // of vowels occupying
        // last indices of the string
        if (cnt != 0)
 
            ans += (cnt * (cnt + 1) / 2);
 
        return ans;
    }
 
    // Driver Program
    static public void main(String[] args)
    {
        String s = "geeksforgeeks";
 
        System.out.println(CountTotal(s));
    }
}

Python3

# Python3 program to Count all substrings
# in a string which contains only vowels
 
# function to check vowel or not
def isvowel(ch):
    return(ch in "aeiou")
 
 
# Function to Count all substrings
# in a string which contains
# only vowels
def CountTotal(s):
     
    ans = 0
    n = len(s)
    cnt = 0
     
    for i in range(0, n):
         
        # if current character is
        # vowel
        if(isvowel(s[i])):
            # increment
            cnt += 1
             
                 
        else:
             
            # Count all possible
            # substring of calculated
            # length
            ans += (cnt*(cnt + 1) // 2)
                 
            # Reset the length
            cnt = 0
 
    # Add remaining possible
    # substrings consisting
    # of vowels occupying
    # last indices of the string
    if(cnt != 0):
        ans += (cnt*(cnt + 1) // 2)
 
    return ans
 
# Driver Program
s = "geeksforgeeks"
 
print(CountTotal(s))

C＃

// C# program to count all substrings
// in a string which contains only vowels
using System;
 
class GFG{
 
// Function to check vowel or not
static bool isvowel(char x)
{
    return (x == 'a' || x == 'e' ||
            x == 'i' || x == 'o' ||
            x == 'u');
}
 
// Function to find largest
// string which satisy condition
static int CountTotal(string str)
{
    int ans = 0;
    int n = str.Length;
    char[] s = str.ToCharArray();
    int cnt = 0;
 
    for(int i = 0; i < n; i++)
    {
        
       // Check if current character
       // is vowel
       if (isvowel(s[i]))
        
           // Increment count
           cnt += 1;
       else
       {
            
           // Count all possible
           // substrings of calculated
           // length
           ans += (cnt * (cnt + 1) / 2);
            
           // Reset the length
           cnt = 0;
       }
    }
     
    // Add remaining possible
    // substrings consisting
    // of vowels occupying
    // last indices of the string
    if (cnt != 0)
        ans += (cnt * (cnt + 1) / 2);
 
    return ans;
}
 
// Driver code
static public void Main(string[] args)
{
    string s = "geeksforgeeks";
 
    Console.Write(CountTotal(s));
}
}
 
// This code is contributed by rock_cool

Java脚本

 
    // Javascript program to count all substrings
    // in a string which contains only vowels
     
    // Function to check vowel or not
    function isvowel(x)
    {
        return (x == 'a' || x == 'e' ||
                x == 'i' || x == 'o' ||
                x == 'u');
    }
 
    // Function to find largest
    // string which satisy condition
    function CountTotal(str)
    {
        let ans = 0;
        let n = str.length;
        let s = str.split('');
        let cnt = 0;
 
        for(let i = 0; i < n; i++)
        {
 
           // Check if current character
           // is vowel
           if (isvowel(s[i]))
 
               // Increment count
               cnt += 1;
           else
           {
 
               // Count all possible
               // substrings of calculated
               // length
               ans += (cnt * (cnt + 1) / 2);
 
               // Reset the length
               cnt = 0;
           }
        }
 
        // Add remaining possible
        // substrings consisting
        // of vowels occupying
        // last indices of the string
        if (cnt != 0)
            ans += (cnt * (cnt + 1) / 2);
 
        return ans;
    }
     
    let s = "geeksforgeeks";
  
    document.write(CountTotal(s));
     

输出：

7

时间复杂度： O(N)